Optimal. Leaf size=279 \[ \frac {(a+b \sec (c+d x))^{n+1} \, _2F_1\left (1,n+1;n+2;\frac {a+b \sec (c+d x)}{a-b}\right )}{2 d (n+1) (a-b)}+\frac {(a+b \sec (c+d x))^{n+1} \, _2F_1\left (1,n+1;n+2;\frac {a+b \sec (c+d x)}{a+b}\right )}{2 d (n+1) (a+b)}-\frac {(a+b \sec (c+d x))^{n+1} \, _2F_1\left (1,n+1;n+2;\frac {b \sec (c+d x)}{a}+1\right )}{a d (n+1)}-\frac {b (a+b \sec (c+d x))^{n+1} \, _2F_1\left (2,n+1;n+2;\frac {a+b \sec (c+d x)}{a-b}\right )}{4 d (n+1) (a-b)^2}+\frac {b (a+b \sec (c+d x))^{n+1} \, _2F_1\left (2,n+1;n+2;\frac {a+b \sec (c+d x)}{a+b}\right )}{4 d (n+1) (a+b)^2} \]
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Rubi [A] time = 0.23, antiderivative size = 279, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 5, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {3885, 961, 68, 65, 831} \[ \frac {(a+b \sec (c+d x))^{n+1} \, _2F_1\left (1,n+1;n+2;\frac {a+b \sec (c+d x)}{a-b}\right )}{2 d (n+1) (a-b)}+\frac {(a+b \sec (c+d x))^{n+1} \, _2F_1\left (1,n+1;n+2;\frac {a+b \sec (c+d x)}{a+b}\right )}{2 d (n+1) (a+b)}-\frac {(a+b \sec (c+d x))^{n+1} \, _2F_1\left (1,n+1;n+2;\frac {b \sec (c+d x)}{a}+1\right )}{a d (n+1)}-\frac {b (a+b \sec (c+d x))^{n+1} \, _2F_1\left (2,n+1;n+2;\frac {a+b \sec (c+d x)}{a-b}\right )}{4 d (n+1) (a-b)^2}+\frac {b (a+b \sec (c+d x))^{n+1} \, _2F_1\left (2,n+1;n+2;\frac {a+b \sec (c+d x)}{a+b}\right )}{4 d (n+1) (a+b)^2} \]
Antiderivative was successfully verified.
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Rule 65
Rule 68
Rule 831
Rule 961
Rule 3885
Rubi steps
\begin {align*} \int \cot ^3(c+d x) (a+b \sec (c+d x))^n \, dx &=\frac {b^4 \operatorname {Subst}\left (\int \frac {(a+x)^n}{x \left (b^2-x^2\right )^2} \, dx,x,b \sec (c+d x)\right )}{d}\\ &=\frac {b^4 \operatorname {Subst}\left (\int \left (\frac {(a+x)^n}{4 b^3 (b-x)^2}+\frac {(a+x)^n}{b^4 x}-\frac {(a+x)^n}{4 b^3 (b+x)^2}-\frac {x (a+x)^n}{b^4 \left (-b^2+x^2\right )}\right ) \, dx,x,b \sec (c+d x)\right )}{d}\\ &=\frac {\operatorname {Subst}\left (\int \frac {(a+x)^n}{x} \, dx,x,b \sec (c+d x)\right )}{d}-\frac {\operatorname {Subst}\left (\int \frac {x (a+x)^n}{-b^2+x^2} \, dx,x,b \sec (c+d x)\right )}{d}+\frac {b \operatorname {Subst}\left (\int \frac {(a+x)^n}{(b-x)^2} \, dx,x,b \sec (c+d x)\right )}{4 d}-\frac {b \operatorname {Subst}\left (\int \frac {(a+x)^n}{(b+x)^2} \, dx,x,b \sec (c+d x)\right )}{4 d}\\ &=-\frac {\, _2F_1\left (1,1+n;2+n;1+\frac {b \sec (c+d x)}{a}\right ) (a+b \sec (c+d x))^{1+n}}{a d (1+n)}-\frac {b \, _2F_1\left (2,1+n;2+n;\frac {a+b \sec (c+d x)}{a-b}\right ) (a+b \sec (c+d x))^{1+n}}{4 (a-b)^2 d (1+n)}+\frac {b \, _2F_1\left (2,1+n;2+n;\frac {a+b \sec (c+d x)}{a+b}\right ) (a+b \sec (c+d x))^{1+n}}{4 (a+b)^2 d (1+n)}-\frac {\operatorname {Subst}\left (\int \left (-\frac {(a+x)^n}{2 (b-x)}+\frac {(a+x)^n}{2 (b+x)}\right ) \, dx,x,b \sec (c+d x)\right )}{d}\\ &=-\frac {\, _2F_1\left (1,1+n;2+n;1+\frac {b \sec (c+d x)}{a}\right ) (a+b \sec (c+d x))^{1+n}}{a d (1+n)}-\frac {b \, _2F_1\left (2,1+n;2+n;\frac {a+b \sec (c+d x)}{a-b}\right ) (a+b \sec (c+d x))^{1+n}}{4 (a-b)^2 d (1+n)}+\frac {b \, _2F_1\left (2,1+n;2+n;\frac {a+b \sec (c+d x)}{a+b}\right ) (a+b \sec (c+d x))^{1+n}}{4 (a+b)^2 d (1+n)}+\frac {\operatorname {Subst}\left (\int \frac {(a+x)^n}{b-x} \, dx,x,b \sec (c+d x)\right )}{2 d}-\frac {\operatorname {Subst}\left (\int \frac {(a+x)^n}{b+x} \, dx,x,b \sec (c+d x)\right )}{2 d}\\ &=\frac {\, _2F_1\left (1,1+n;2+n;\frac {a+b \sec (c+d x)}{a-b}\right ) (a+b \sec (c+d x))^{1+n}}{2 (a-b) d (1+n)}+\frac {\, _2F_1\left (1,1+n;2+n;\frac {a+b \sec (c+d x)}{a+b}\right ) (a+b \sec (c+d x))^{1+n}}{2 (a+b) d (1+n)}-\frac {\, _2F_1\left (1,1+n;2+n;1+\frac {b \sec (c+d x)}{a}\right ) (a+b \sec (c+d x))^{1+n}}{a d (1+n)}-\frac {b \, _2F_1\left (2,1+n;2+n;\frac {a+b \sec (c+d x)}{a-b}\right ) (a+b \sec (c+d x))^{1+n}}{4 (a-b)^2 d (1+n)}+\frac {b \, _2F_1\left (2,1+n;2+n;\frac {a+b \sec (c+d x)}{a+b}\right ) (a+b \sec (c+d x))^{1+n}}{4 (a+b)^2 d (1+n)}\\ \end {align*}
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Mathematica [A] time = 7.01, size = 256, normalized size = 0.92 \[ \frac {\cot ^2(c+d x) \left (a+b \sqrt {\sec ^2(c+d x)}\right ) (a+b \sec (c+d x))^n \left ((a-b) \left (a (a-b) (2 a-b (n-2)) \tan ^2(c+d x) \, _2F_1\left (1,n+1;n+2;\frac {a+b \sqrt {\sec ^2(c+d x)}}{a+b}\right )-2 (a+b) \left (2 \left (a^2-b^2\right ) \tan ^2(c+d x) \, _2F_1\left (1,n+1;n+2;\frac {\sqrt {\sec ^2(c+d x)} b}{a}+1\right )+a (n+1) \left (a-b \sqrt {\sec ^2(c+d x)}\right )\right )\right )+a (a+b)^2 (2 a+b (n-2)) \tan ^2(c+d x) \, _2F_1\left (1,n+1;n+2;\frac {a+b \sqrt {\sec ^2(c+d x)}}{a-b}\right )\right )}{4 a d (n+1) (a-b)^2 (a+b)^2} \]
Warning: Unable to verify antiderivative.
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fricas [F] time = 0.74, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (b \sec \left (d x + c\right ) + a\right )}^{n} \cot \left (d x + c\right )^{3}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \sec \left (d x + c\right ) + a\right )}^{n} \cot \left (d x + c\right )^{3}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 1.13, size = 0, normalized size = 0.00 \[ \int \left (\cot ^{3}\left (d x +c \right )\right ) \left (a +b \sec \left (d x +c \right )\right )^{n}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \sec \left (d x + c\right ) + a\right )}^{n} \cot \left (d x + c\right )^{3}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int {\mathrm {cot}\left (c+d\,x\right )}^3\,{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^n \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a + b \sec {\left (c + d x \right )}\right )^{n} \cot ^{3}{\left (c + d x \right )}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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